Title Straight Line Engineering Practice Sheet Solution 25.pdf ✅

Rhombus Publications WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. GKwU mij‡iLvi x I y Aÿ Øviv LwÐZ AskØq abvZ¥K Ges g~jwe›`y †_‡K Aw1⁄4Z j¤^, mij‡iLv‡K mgwØLwÐZ K‡i| j‡¤^i •`N ̈© 7 GKK n‡j, mij‡iLvi mgxKiY wbY©q Ki| [BUET 23-24] mgvavb: awi, mij‡iLvi mgxKiY: xcos + ysin = p GLv‡b, p = 7 †h‡nZz, g~jwe›`y †_‡K Aw1⁄4Z j¤^, mij‡iLv‡K mgwØLwÐZ K‡i|   = 45 Y X 7 O  xcos45 + ysin45 = 7  x 2 + y 2 = 7  x + y – 7 2 = 0 (Ans.) 2. (k, – 3k), (5, k) Ges (– k, 2) we›`yÎq Øviv Drcbœ wÎfz‡Ri †ÿÎdj 28 eM©GKK, †hLv‡b k GKwU c~Y© msL ̈v| wÎfz‡Ri j¤^we›`yi ̄’vbv1⁄4 wbY©q Ki| [BUET 22-23] mgvavb: (k, – 3k), (5, k) Ges (– k, 2) cÖkœg‡Z, 1 2     k – 3k 5 k – k 2 k – 3k = 28  k 2 + 10 + 3k2 + 15k + k2 – 2k = 56  5k2 + 13k – 46 = 0  k = 2, – 23 5 wKš‘ (k  Z)  k = 2  ̄’vbv1⁄4 A(2, – 6); B(5, 2); C(– 2, 2) B C A O(x, y) awi, j¤^we›`y O(x, y)| AB I OC ci ̄úi j¤^ nIqvq, mAB  mOC = – 1  2 + 6 5 – 2  y – 2 x + 2 = – 1  3x + 8y = 10 .....(i) Avevi, AC I OB ci ̄úi j¤^ nIqvq, mAC  mOB = – 1  2 + 6 – 2 – 2  y – 2 x – 5 = – 1  x – 2y = 1 ....(ii) (i) I (ii) mgvavb K‡i, wb‡Y©q j¤^we›`y(x, y)      2  1 2 (Ans.) 3. (a – 1)x + y = a ........ (i) (2, 2)  (b, 0) ........... (ii) †iLvØq (K) j¤^ n‡j a I b Gi m¤úK© Kx? (L) mgvšÍivj n‡j a I b Gi m¤úK© Kx? [BUET 21-22] mgvavb: †`Iqv Av‡Q, (a – 1)x + y = a  y = – (a – 1)x + a GB †iLvi Xvj, m1 = – (a – 1) Ges (2, 2) Ges (b, 0) we›`y؇qi ms‡hvMKvix †iLvi Xvj m2 = 2 – 0 2 – b = 2 2 – b (K) †iLvØq j¤^ n‡j, m1m2 = – 1  – (a – 1) × 2 2 – b = – 1  2a – 2 = 2 – b  2a + b = 4 (Ans.) (L) †iLvØq mgvšÍivj n‡j, m1 = m2  – (a – 1) = 2 2 – b  (a – 1)(2 – b) = – 2  2a – ab – 2 + b = – 2  2a + b – ab = 0 (Ans.) 4. g~j we›`y n‡Z xsec – ycosec = k Ges xcos – ysin = kcos2 †iLv؇qi j¤^ `~iZ¡ h_vμ‡g 2 cm Ges 3 cm| k Gi gvb wbY©q Ki| [BUET 19-20] mgvavb: g~jwe›`y (0, 0) †_‡K xsec – ycosec – k = 0 †iLvi Dci j¤^`~iZ¡,       0 × sec – 0 × cosec – k sec2  + cosec2  = 2  k 2 1 cos2  + 1 sin2  = 4 [eM© K‡i]


4  Higher Math 1st Paper Chapter-3 Rhombus Publications awi, C (x, y) fi‡K‡›`ai m~Î n‡Z, x + 1 – 1 3 = 0  x = 0 y + 0 + 0 3 = 1 3  y = 3  AD †iLvi Xvj = 1 3 – 0 0 + 1 = 1 3  BC †iLvi Xvj = 0 – 3 0 + 1 = – 3  AD †iLvi Xvj × BC †iLvi Xvj = 1 3 × ( – 3) = – 1  AD  BC (Showed) weMZ mv‡j KUET-G Avmv cÖkœvejx 16. hw` 3x + by + 1 = 0 Ges ax + 6y + 1 = 0 mij‡iLvØq (5, 4) we›`y‡Z †Q` K‡i, Z‡e a Ges b Gi gvb wbY©q Ki| hw` cÖ_g †iLvwU x Aÿ‡K A we›`y‡Z Ges wØZxq †iLvwU y Aÿ‡K B we›`y‡Z †Q` K‡i, Z‡e AB mij‡iLvi mgxKiY wbY©q Ki| [KUET 19-20] mgvavb: †iLvØq (5, 4) we›`yMvgx e‡j, 3  5 + b.4 + 1 = 0  b = – 4 a.5 + 6  4 + 1 = 0  a = – 5 cÖ_g †iLv: 3x – 4y + 1 = 0  x – 1 3 + y 1 4 = 1  A     – 1 3  0 wØZxq †iLv: – 5x + 6y + 1 = 0  x 1 5 + y – 1 6 = 1  B     0  – 1 6 AB Gi mgxKiY, y – 0 = – 1 6 – 0 0 + 1 3     x + 1 3  – 6y = 3x + 1  3x + 6y + 1 = 0 (Ans.) 17. LM mij‡iLvwU g~jwe›`y n‡Z 5 GKK `~ieZ©x Ges x I y Aÿ‡K h_vμ‡g A I B we›`y‡Z †Q` K‡i| g~jwe›`y n‡Z LM Gi Dci Aw1⁄4Z j¤^ y A‡ÿi †hvM‡evaK w`‡Ki mv‡_  3 †KvY Drcbœ K‡i| C we›`yi ̄’vbv1⁄4 (– 1, – 2) n‡j, ABC wÎfz‡Ri †ÿÎdj wbY©q Ki| [KUET 19-20] mgvavb: Y Y X X LM LM B 5 5  3  3 C(– 1, – 2) A A  LM 1g PZzf©v‡M Dfq Aÿ‡K †Q` Ki‡j,  = 90 – 60 = 30 j‡¤^i •`N© ̈ p = 5 GKK  LM  xcos + ysin = p  xcos30 + ysin30 = 5  x 10 3 + y 10 = 1  A     10 3  0 , B(0, 10), C(– 1, – 2)  ABC = 1 2       10 3 0 – 1 0 10 – 2 1 1 1 = 5 + 20 3 eM© GKK (Ans.) LM 2q PZzf©v‡M Dfq Aÿ‡K †Q` Ki‡j,  = 90 + 60 = 150  LM  xcos150 + ysin150 = p  x – 10 3 + y 10 = 1  A    –  10 3  0 , B(0, 10), C(– 1, – 2)  ABC = 1 2      –  10 3 0 – 1 0 10 – 2 1 1 1 = 20 3 – 5 eM© GKK (Ans.) 18. 3x + 4y = 11 Ges 12x – 5y = 2 †iLv؇qi AšÍfz©3 m~2‡Kv‡Yi mgwØLЇKi mgxKiY wbY©q Ki| [KUET 13-14, 06-07; CUET 13-14, 07-08; BUET 06-07] mgvavb: wb‡Y©q mgwØLÐK, 3x + 4y – 11 3 2 + 42 =  12x – 5y – 2 122 + 52 †h‡nZz, a1a2 + b1b2 > 0  (– ve) wb‡q m~2‡Kv‡Yi mgwØLÐK cvIqv hv‡e|  3x + 4y – 11 5 = – 12x – 5y – 2 13  39x + 52y – 143 = – 60x + 25y + 10  99x + 27y – 153 = 0  11x + 3y – 17 = 0 (Ans.)